Difference between revisions of "2021 Fall AMC 12B Problems/Problem 7"

Revision as of 10:34, 24 November 2021

The following problem is from both the 2021 Fall AMC 10B #12 and 2021 Fall AMC 12B #7, so both problems redirect to this page.

Which of the following conditions is sufficient to guarantee that integers $x$ , $y$ , and $z$ satisfy the equation $x(x-y)+y(y-z)+z(z-x) = 1?$

$\textbf{(A)} \: x>y$ and $y=z$

$\textbf{(B)} \: x=y-1$ and $y=z-1$

$\textbf{(C)} \: x=z+1$ and $y=x+1$

$\textbf{(D)} \: x=z$ and $y-1=x$

$\textbf{(E)} \: x+y+z=1$

Plugging in every choice, we see that choice $\textbf{(D)}$ works.

We have $y=x+1, z=x$ , so $x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.$ Our answer is $\textbf{(D)}$ .

~kingofpineapplz

Just plug in all these options one by one, and one sees that all but $D$ fails to satisfy the equation.

For $D$ , substitute $z=x$ and $y=x+1$ :

$LHS=x(x-(x+1))+(x+1)(x+1-x)+x(x-x)=(-x)+(x+1)=1=RHS$

Hence the answer is $\boxed{\textbf{(D)}}.$

~Wilhelm Z

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 13: / Line 13: @@
 <math>\textbf{(E)} \: x+y+z=1</math>
+==Solution 1==
+Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.
+We have <math>y=x+1, z=x</math>, so
+<cmath>x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.</cmath>
+Our answer is <math>\textbf{(D)}</math>.
+~kingofpineapplz
-==Solution 1 (Bash) ==
+==Solution 2 (Bash) ==
 Just plug in all these options one by one, and one sees that all but <math>D</math> fails to satisfy the equation.
@@ Line 26: / Line 35: @@
 ~Wilhelm Z
-==Solution 2==
-Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.
-We have <math>y=x+1, z=x</math>, so
-<cmath>x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.</cmath>
-Our answer is <math>\textbf{(D)}</math>.
-~kingofpineapplz
 {{AMC12 box|year=2021 Fall|ab=B|num-a=8|num-b=6}}
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